Dan Christensen <Dan_Christensen@sympatico.ca> wrote:> On May 1, 6:56=A0pm, David W. Cantrell <DWCantr...@sigmaxi.net> wrote:> > "Jesse F. Hughes" <je...@phiwumbda.org> wrote:> >> > > I wonder if you [Dan] can find a calculus text which explicitly> > > leaves 0^0 undefined.> >> > I have several introductory calculus texts on the shelves.> >> > Some authors feel the need to review pre-calculus topics. Thomas &> > Finney, 7th ed., do this in an appendix. They say that, if a is> > nonzero, a^0 = 1.> > But of course, they give no reason for excluding a = 0.> >> > But unless the author feels the need to review pre-calculus topics,> > there may be no mention of 0^0 until the section on power series.> > Perhaps Leithold, 5th ed., is typical. [Thomas & Finney do something> > similar regarding power series.] Soon after defining power series,> > Leithold states> >> > "(Note that we take (x - a)^0 =3D 1, even when x =3D a, for convenience> > in writing the general term.)"> >> > I had more-or-less expected to find such a statement, but was surprised> > by a similarly phrased statement on the next page:> >> > "When n! is used in representing the nth term of a power series, we> > take 0! = 1 so that the expression for the nth term will hold when n = 0.">> Again, I have no trouble with 0! = 1

Of course, 0! = 1. But to be consistent, since both 0^0 and 0! are emptyproducts, you should either accept or reject that both are 1. (Of course,you and Leithold _should_ accept that both are 1.)

From what I see below, you intend "natural number" to include 0. That'sfine; I normally do that myself.

But is that a _definition_ of the factorial? We put the definiendum (thething _being_ defined) on the left and the definiens on the right. If youhave done that, then your first line defines 1! to be 1. But then yoursecond line, when x = 0, attempts to REdefine 1! in terms of somethingwhich has yet to be defined, namely, 0!. Surely you see that that is aproblem!

> No need for any empty products.>> We could then legitimately define the ! function as:>> x! = {1 for x = 0> {(x-1)! * x for x > 0

"then"? It wouldn't have been illegitimate even before then.

But thinking of 0! as the empty product provides motivation for itsdefinition.

For natural n (including 0), I like to think of n! as being simply

| The product of the first n positive integers.

And then naturally this gives us 0! = 1.

David W. Cantrell

> > Wow! Horrible! But at least Leithold is consistent: Both 0^0 and 0! are> > empty products and he treats them in essentially same way, making it> > seem that they aren't _really_ 1, but will just be taken to be 1 in the> > context of power series, for convenience. Aaargh.>> Interesting...>> Dan> Download my DC Proof 2.0 software at http://www.dcproof.com> Also see video demo